# DC Mathematica 2017

0 2017

DULWICH COLLEGE | MATHEMATICA 2017

Editors:

Theo Macklin

Phillip Cloud

Supervisor:

Dr Purchase

Cover Art By:

Theo Podger

Contributors:

Simon Xu

Jakub Dranczewski

Andrew Ng

Hin Chi Lee

Timothy Moulding

Toby Evans

Jay Connor

Lunzhi Shi

Joshua du Parc Braham

Joseph Lazzaro | Ayman D’Souza

Jack Kurtulus

Theo Macklin

Mr Ottewill

1

CONTENTS

Interesting Integer Sequences and Their Stories

3

Lunzhi Shi

Y10

Can You Trisect an Angle?

8

Simon Xu

Y11

Measuring Infinities: Or a short journey into the number theory

11

Jakub Dranczewski

Y13

14

Toby Evans

Y9

Parity

17

Hin Chi Lee

Y11

Taylor Series and Euler’s identity

21

Andrew Ng

Y12

What is the Probability of Two Random Numbers Being Coprime?

24

Theo Macklin Y12

Are the Renewed Statistics and Probability Behind the "Draft Lottery" Used in the NBA Reliable and an Accurate Method of Levelling a Team's Ability? 28 Jack Kurtulus Y13

A Golden Opportunity: Mathematical Patterns in Nature

33

Jay Connor

Y9

Sequences and Architecture

35

Timothy Moulding

Y10

Modelling Infectious Diseases

42

Joseph Lazzaro & Ayman D’Souza

Y12

Fractals: Are They Just Mathematical Curiosities?

46

Joshua du Parc Braham Y13

From a Question by Dr Purchase

55

Mr Ottewill

Puzzles

60

2

Interesting Integer Sequences and Their Stories

Lunzhi Shi

I was first introduced to the Pascal’s Triangle back in year 5 when I was asked to produce a program that loops and gives a constant output of the triangle. As I began to learn more mathematics, I realized that the triangle was far more than an array as you see it.

If you aren’t sure about what Pascal’s Triangle is, it’s like the food triangle: the further down the more there is. One number in the array is simply the sum of the two above it, taking empty spaces as zero and the initial as 1: it keeps on going non-stop. Now over with the definition, let’s first take a look at the applications of the triangle. I will start off with this easy one related to everyday algebra: Try expanding this: ( + ) . Now this: ( + ) 2 . That will be 2 + 2 +  2 . How about ( + ) 3 ? It will get more and more complicated.

But why am I mentioning binomial expansion? Is that somehow related to the triangle? Have a look at the rearranged table of the expansions (up to the power of 5):

Row Expansion 0

1 +

1 2 3 4 5

1 x

1 y

1 x 2

1 y 2

+

2 xy

+

1 x 3

3 x 2 y

3 xy 2

1 y 3

+

+

+

1 x 4

4 x 3 y

6 x 2 y 2

4 xy 3

1 y 4

+

+

+

+

1 x 5

5 x 4 y

10 x 3 y 2

10 x 2 y 3

5 xy 4

1 y 5

+

+

+

+

+

As you can see, the coefficients draw out the first five rows of the Pascal’s Triangle. Say we have ( + )  , you then need to find row  to get the coefficient. The row and the expansion are magically related. The powers also come in with a very clear pattern of adding up and subtracting down to zero. This fun fact leads to so called binomial theorem. To further explain this, you possibly know that we need to use (  ) or ( choose  ). This is combinatorics; representing the number of ways of picking 

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elements from a set of elements. We can deduce the numeric value of ( choose  ) using the formula:



! ! ( − )!

(

) =

Now have a look at the pattern to the right – we have formed Pascal’s Triangle. Using this method, we can create a bridge to better understand the relationship between Binomial Theorem and Pascal’s Triangle.

Therefore, according to what’s shown above, for (x + y) n , we can write down an expansion of:

 0

 1

 2 ) −2  2 + ⋯+ (

  − 1

 

( + )  = (

)   0 + (

) −1  1 + (

) 1  −1 + (

) 0  

Or:

( + )  = ∑(  𝑘 )  =0

−  

hence binomial theorem.

If we expand the binomial theorem further into multinomial theorem it’s going to become more interesting. We can first have a look at the example ( +  + )  . As it is three- dimensional, we can visualize this with a matching Pascal’s Pyramid. And if we go into factorizing, it can be represented as:

( +  + )  = ∑     −−  ,

What if we have, say, subjects in the brackets and visualizing polygons can no longer support us? Since everything follows a pattern, there is a formula for multinomial theorem:





)  = ∑ (

( 1

+ 2

+ ⋯+

)∏ 

𝑘 1

, 𝑘 2

, … , 𝑘

 1 + 2 +⋯+ =

=1

Going through the full proof would take a very long time. There are articles you can find online that explain the theorem step by step – it only adds

another loop of getting a product on top of the coefficient calculation.

Now, on to some other sequences, there are a lot to find in this triangle and I will go through some briefly. Say we take all the odd numbers in a Pascal’s Triangle and leave the rest

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alone. Zoom out, we discover this pattern known as the Sierpinski Triangle. Same goes for the pyramid.

We also have the Fibonacci sequence which is a bit hard to spot.

You have possibly worked with probabilities with this sequence in school. A famous example is flipping coins. You will either get heads or tails. Suppose you throw it  times. The possible number of results can be found in row  in Pascal’s triangle. The total possibilities of either heads or tails appearing times, ( ≤ ) , is shown in the array with the ℎ representing the total possibilities. Let’s draw a table:

Throws

Possible Results

Pascal’s Triangle

Throws

Possible Results

Pascal’s Pyramid

H T S

1 1 1 1 2 1 2 1 2

H T

1 1 1 2 1 1 3 3 1

1

1

HH HT TH TT

HH HT TH TT TS ST SS SH HS

2

2

HHH HHT HTH THH HTT THT TTH TTT

3

The same theory applies to the pyramid model. Suppose a very skilled coin flipper has 1 3 the probability of landing a coin standing straight up, the table would look like the one above on the right. So where does this perfect creation come from? We need to look back to France in 1654. In 1654 Blaise Pascal completed his Traité du Triangle Arithmétique and later the titular array was given the name Pascal’s Triangle. However, the original discovery of this array, which later led to the more famous Pascal’s Triangle, actually took place in China. In the 13 th century, mathematician Yang Hui officially presented the triangle. Hence it is still called the Yang Hui Triangle, 杨辉 三角 , in China. The earliest official mathematics in China dates back to the Shang Dynasty (1600-1050 BC). Back then, calculations were done using rods of bamboo. This first number system could even handle complex calculations pretty well. It was, however, just a counting method- the written numbers were not that efficient. Jumping 1500 years ahead, the numeric system had improved a lot to make mathematics more accessible. In the 13 th century, China’s mathematics reached its peak as many new features were discovered by mathematicians and the government decided to make education of mathematics to everyone possible.

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Yang Hui was one of the mathematicians that stood out during this time. To the right is the “Yang Hui’s Magic Concentric Circle”, another very interesting integer array. Counting the four circles and the eight semi-circles, the sum of the numbers in each semi- circle is exactly 69. There are also eight radii which, if you add the numbers up, also give 69. The second sum

is 147 – which is the sum of the numbers on four diameters and the sum of the numbers in each circle plus 9 in the center.

Talking about magic circles, the Ding Yi Dong ( 丁易 东 ) magic circle is also worth mentioning. In this model, the sum of each circle is 200. Also, taking a number on one circle and adding it to the number which is exactly on the opposite side of that circle, you always get 50 (for example, 14+36=50 and 45+5=50). Knowing some of these properties, we can deduce other properties like the sum of each diameter would always be 325 (25+6x50).

Magic squares, in turn, are constructed with similar properties. Again, the square is always related to a ‘magic’ constant which is the sum of the integers in each row, column and diagonal. In this 3 x 3 model, it is 15. This model is called the Lo Shu Square, which is the earliest magic square discovered in China. The more advanced magic squares were officially recorded in Greece in the 14 th century. These models were combined with a much

deeper meaning- linked to planets and gods. The magic squares were said to be written into medieval magic books. Ranging from the 3 x 3 model to the 9 x 9 one, they were given names of: Saturn, Jupiter, Mars, Sol (Sun), Venus, Mercury, Luna (Moon). Astronomy is very different though, but still very well connected with maths. Later people constructed the magic cube that has similar properties. Well, people never stop exploring, do they? Soon the term ‘magic hypercubes’ were introduced by mathematician John R. Hendricks.

The cubes follow the exact same rule, that is, the sum on any axis is always the same. Let’s have a look at the previously innocent-looking 3x3 model in 4d. People have been creating crazy things all the time, including the ones they can’t really see, or understand very well.

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You might wonder the formula for the magic number, since no matter how you construct this magic cube, the stem of the magic number will be related to it. Suppose a n -dimensional magic cube arranged in an  ×  × …×  structure, the formula for the magic number would be:

(  + 1) 2

() =

 

Looking into the higher dimensions has always been a hard thing to do, and the seemingly convincing results we get could sometimes be more confusing than the question itself. Imagine throwing a paper ball to a whiteboard with 2-dimensional creatures on it. What they would see would only be a cross-section of the

paper ball and its changing shapes will not follow their laws of physics. But just now, we put ourselves as ‘flatlanders’. The 4d model above is technically correct, but if a 4d visitor decides to take you into his dimension to show you this magic cube, you would be seeing blobs and spheres and other things that make absolutely no sense - could this explain why movie makers decide to put a blurring and confusing background every time there’s a scene of time travel? There are so many integer sequences out there and you can just pick one from the pool and spend the whole afternoon looking into it. There is even an encyclopaedia just for integer sequences – oeis.org , or The Online Encyclopaedia of Integer Sequences. By 2015, there were at least 250,000 sequences on their list – go have a look and see if you find something new. The earliest, pre-historic mathematics that could be 20,000 years old started as human began to quantify things around them – time, matter, etc. The concept of numbers is always the essence of our world and writing this article has been a good experience; reminding me of a page I haven’t turned over for a long time. And most importantly of all, I hope you, fellow mathematicians, have enjoyed reading this.

7

Can You Trisect an Angle in a Straight Edge Compass Construction?

Simon Xu

Using a straight edge and a compass we can construct many lengths if we are given a unit length. If we take the unit length as ‘one’, we can copy its length using the compass and add it to itself to construct ‘two’; add another one to construct ‘three’; take one away from ‘four’ to make ‘three’. When we can do addition and subtraction, multiplication and division a length can be made as well to represent the answer. Here we can see how some geometrical operations are equivalent to axiomatic algebra.

So what are the rules for straightedge-compass construction? What is the limit for what number we can construct?

Rules for straight/edge compass construction:

Basic operations for straight- edge/ compass construction: Creating the length through two existing points Creating a circle through one point with centre another point Creating the point which is the intersection of two existing, non-parallel lines Creating the one or two points in the intersection of a line and a circle (if they intersect) Creating the one or two points in the intersection of two circles (if they intersect)     

In terms of algebra, a length is constructible if and only if it represents a constructible number, and an angle is constructible if and only if its cosine is a constructible number. All rational numbers are constructible, and all constructible numbers are algebraic numbers. A field of rational numbers ( ℚ ) is a set containing operations including addition and multiplication. Also, if and  are constructible numbers with  ≠ 0 , then ±  , ×  , , and √ are all constructible. A complex number is constructible if, and only if, the real and imaginary parts are both constructible. In a straight-edge/compass construction, we can also create numbers such as root 2 by using the method as shown on the right, where we can construct as any rational number, and we can construct a number = + √ , where ,  ,  are rational numbers, which forms a new field, F1, based on the field of rational numbers, 𝐹0 . We can construct another number from another field by constructing  = + √ where ,  ,  are from 𝐹1 , and so on we can construct many numbers and values. angle, and then prove that one third of that angle is not constructible, so we use that as an example to prove that we cannot trisect that angle. To do this, we need to prove that we cannot construct a number such as (40) , which means that an angle of 40˚ is not constructible, but as we can construct a 120˚ angle and therefore the operation of trisecting a 120˚ angle is not possible.  Therefore, to prove that a given angle cannot be trisected in straight-edge/compass construction, we need to construct an

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Theorem: If a cubic has any constructible root then it has at least one real root.

Proof:

Suppose that the in cubic:

3 +  2 +  +  = 0

there are constructible roots but none in ℚ (the set of rational numbers) and there is some value such that the field 𝐹() is the closest field to the rational number field in which a root exists. I.e. that the root is = + √𝑘 for ,  , 𝑘 from a field lower than 𝐹() , 𝐹( − 1) and √𝑘 is in 𝐹() .

Then we put = + √𝑘 into the cubic:

We can create many fields from the field of rational numbers

( + √𝑘) 3 + ( + √𝑘) 2 + ( + √𝑘) +  = 0

If we expand the brackets we get:

( + 3 +  2 𝑘 +  2 +  2 𝑘 + ) + (3 2  +  3 𝑘 + 2 + )√𝑘 = 0

which we can simplify as :

 + √𝑘 = 0

So either √𝑘 = −/ or  =  = 0 . Because −/ is in 𝐹( − 1) and √𝑘 is not, this case cannot be true; therefore S=T=0 must be the true case.

Therefore we ask: is = − √𝑘 a root?

( − √𝑘) 3 + ( − √𝑘) 2 + ( − √𝑘) +  = 0

( + 3 +  2 𝑘 +  2 +  2 𝑘 + ) − (3 2  +  3 𝑘 + 2 + )√𝑘 = 0

∴  − √𝑘 = 0

Because  =  = 0 , this is also true. However, a cubic cannot have just 2 real roots (unless the roots are repeated). Let +√𝑘 = 1 , − √𝑘 = 2 , and the third (either repeated or not repeated) be 3 . I.e.:

( − 1

)( − 2

)( − 3

) = 0

3 + ( 1

) 2 + ⋯ = 0

+ 2

+ 3

Comparing with the original cubic, 3 +  2 +  +  = 0 , we can say that:

) = −( + + 3

) = −2 − 3

 = ( 1

+ 2

+ 3

3

= −2 − 

9

Because a and p are in the field 𝐹( − 1) , 3 is in the field 𝐹( − 1) . This contradicts with what we supposed at the beginning in which 𝐹() is the lowest field which has a constructible root. So if a cubic has constructible roots, at least one root is rational; if a cubic has no rational roots, it has no constructible roots.

How can we use this theorem?

To prove that we cannot trisect angle with a straight-edge and a compass, we need to show that there exists an angle that cannot be trisected using a straightedge and a compass. For example, a 120˚ angle cannot be trisected, because we can prove that (40) is not constructible. To do this we need to put (40) into a cubic function.

Using the trigonometric identity:

(3) = 4 3 () − 3()

in which =40˚:

(120) = 0.5

0.5 = 4 3 (40) − 3(40)

If we let (40) = then the cubic will become:

8 3 − 6 − 1 = 0

Which we can prove has no rational roots by typing it into a calculator. So we can say that it has no constructible roots. Because (40) is a root of this function, (40) is not constructible, therefore an angle of 120˚ which we can construct cannot be trisected.

In conclusion, we cannot trisect an angle in a straightedge-compass construction.

10

Measuring Infinities: a Short Journey into Number Theory

Jakub Dranczewski

Could you say that one infinity is somehow bigger than another one? It seems completely senseless. How can one claim that one thing of infinite size is greater than another thing of infinite size if, by definition, they both have no proper end? Let’s just stop here for a second though. What does it actually mean that two sets of things have the same number of elements? Your first answer would probably be ‘if you count the elements in them, you end up with the same number’. But what if we cannot count the number of elements in the set? This definition fails in that case. We have to look for something more general, and that something is a simple action: matching the elements of the two compared sets into pairs. If we manage to find a method of matching that leaves no elements of either set without a pair, we have proven that the number of elements in both sets is the same! Is this in any way useful for you? Well, that surely depends on your definition of usefulness, but one can make some fairly decent proofs with the above method. For example, we can quickly prove that the number of even numbers is exactly the same as the number of positive natural numbers! ‘But wait!’, you scream, ‘shouldn’t there be twice as many natural numbers as even numbers? Surely there is an additional odd number for every even number’. Intuitively that is correct, but let’s use the pairing method and see what happens. Take every natural positive number and multiply it by two, then assign the result as a pair to that number. If you write that down you’ll notice that all the even numbers were assigned to a natural positive number. … This means that the number of positive integers and the number of even integers are exactly the same. We just compared two infinite sets! It can be proven in a similar way that the number of integers is the same as the number of negative integers (I for one was initially rather surprised by that fact) or that there are exactly as many points on a line segment as on the entire line. The proofs go on and on! But hey, one could argue that all infinities are of the same size – as they are all infinite this seems to be a fair assumption. So now the real fun begins: let us prove that some infinities are bigger than others! 1 2 3 4 5 6 7 … ▼ ▼ ▼ ▼ ▼ ▼ 10 ▼ 12 ▼ 14 2 4 6 8

Not all infinities are equal

The proof above involved the size of a set of all integers. Let’s call this amount ℵ 0 (aleph-zero, the term is actually used by mathematicians, and it originates from a Hebrew letter). Now to help us find an infinity bigger than that I’ll quickly go through the proof that the size of the set of all rational numbers is also ℵ 0 (bear with me, it’s simple). First, make a table as below:

0 1

1

2

3

4

… … … … … …

1/1

2/1

3/1

4/1

-1

-1/1

-2/1

-3/1

-4/1

2

1/2

2/2

3/2

4/2

-2

-1/2

-2/2

-3/2

-4/2

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Now follow the blue line, omitting doubles, and assign subsequent positive integers to every cell you visit. You just numbered all the rational numbers! Since they are numbered, they all have a pair in the natural numbers set and therefore the amount of them is the same (another name for ℵ 0 -sized sets stems from that proof method: we call them countable since you can count their elements using natural numbers). Thus, the number of rational numbers can also be written as ℵ 0 . Now, onto the greater infinities. Imagine a number line and two rational numbers as close to each other as possible. Notice that there are still a lot of numbers between them that are not rational. As you get closer and closer you discover an infinite amount of real numbers between each pair of rational numbers, which there is, of course, an infinite amount of. It’s like infinity squared! This exclamation is, in fact, a rather close description of the number of real numbers. There are a lot of them, even more than rational numbers. The proof above relayed strongly on intuition, but there is, of course, a more rigorous one, which I will not write down here, but it involves imagining a list of all the rational numbers and creating new numbers out of them that are not on the list. The amount of real numbers is called continuum. This brings us to one of the most interesting problems of the set theory: the continuum hypothesis. The hypothesis simply asks whether there is an infinity that is bigger than ℵ 0 , but smaller than continuum. And yet it landed on David Hilbert’s list of the twenty-two important open questions in mathematics, and generally puzzled mathematicians for a very long time. The interesting part of the hypothesis is that it cannot be proven. It cannot be disproven either. It was mathematically proved to be unprovable, and then mathematically proved to be undisprovable, which is just weird. It just hangs somewhere in-between, neither true nor false, and mathematicians try to come up with new ways to crack it (since it is not impossible that with completely new tools the hypothesis could be one day settled). But we do not need to consider problems this complicated to have some fun with infinities. There are many problems that are easier and still fun. Imagine a hotel that has an infinite amount of numbered rooms, each housing a mathematician. Now, what shall the hotel’s management team do if a new mathematician arrives? They certainly should not just dismiss him or make him sleep on a random couch – mathematicians are sensitive beings; the poor guy could catch a cold or something. The hotel’s staff needs to find an empty room in a hotel in which every single room is already taken. The resolution is actually surprisingly simple – just make every mathematician move to the room next door (if he was in room one, he should go to room two, if he was in room two, he should go to room three, and so on). Now room one is free! A swift and mind-bending solution, involving merely the pain of infinite mathematicians having to change rooms all of a sudden, which they would surely be willing to endure for a colleague in need. … Now what happens if a coach with infinite mathematicians arrives and they all want a room? What if an infinite number of coaches arrive, each with an infinite number of mathematicians inside? The world of infinity problems seems to be truly endless and I do encourage you to look yourself for some of the wonderful and often weird things that happen in there. The Hilbert’s Hotel – the absolute classic of number theory 1 2 3 4 5 6 Is there something in between?

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Sources:

 Most of this article is based on notes taken by me during a lecture by Krzysztof Ciesielski (Ph.D.) at the Jagiellonian University in Krakve. Nevertheless, below I cite some interesting webpages that can be a good start in your number theory research:  https://www.ias.edu/ideas/2011/kennedy-continuum-hypothesis - Can the Continuum Hypothesis Be Solved? by Juliette Kennedy  https://en.wikipedia.org/wiki/Hilbert’s_paradox_of_the_Grand_Hotel - a thorough description and analysis of the Hilbert’s Hotel problem.

13

Toby Evans

Zeno was an Italian Mathematician, he was born in about 490 BC and died around 425 BC but during his brief life he made some great discoveries. Since he lived so long ago, we know little about him; however we do know that he was a philosopher. Many of his paradoxes are known to us today, and in this article, I will explain the paradox of Achilles and the Tortoise. Achilles was a legendary Greek hero; he was a terrific fighter and could run very quickly, whereas a tortoise is an animal known for being slow.

Zeno

Zeno stated that if the tortoise was given a small head start, then despite Achilles being so much faster, he could never catch up and overtake. The logic behind this was that Achilles would have to cover the distance the tortoise had already moved to catch up with it, however by this time the tortoise would have moved on. Hence Achilles would then have to cover that additional distance too, and, whilst doing this, the tortoise would move on even further. This cycle would continue endlessly, with the tortoise always moving further away. Whilst the gap might narrow, the tortoise would always win this race.

The conclusion is that anything, given a head start and no matter how slow it is going, will inevitably beat a faster object, for example a formula one car would lose to a snail if the snail starts the race

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first. This, however, seems unlikely. Therefore, I shall try to disprove Zeno’s statement with an example, using proof by contradiction.

If Achilles is travelling at twice the speed of the tortoise and the tortoise travels 1 metre a second and has a 1 second head start, then the following sequence occurs.

Time /seconds

Achilles /metres

Tortoise /metres

1 2 3 4 5

0 2 4 6 8

1 2 3 4 5

9

Tortoises are slow!

8

7

6

5

4

3

2

1

0

0

1

2

3

4

5

6

Achilles (metres)

Tortoise (metres)

Graph showing the data in the table above

Achilles, therefore, catches up with the tortoise at 2m and overtakes at the 3 second point, he is travelling at double the speed, so makes up the distance quickly.

I will now demonstrate Zeno’s paradox with the same example, but using a different set up for the problem. The start point for each successive time interval,  , is the time at which Achilles has caught up to the point where the tortoise was at t-1.

Time (seconds)

Achilles (metres)

Tortoise (metres)

1

0 1

1

1.5

1.5

1.75

1.5

1.75

1.875

1.75

1.875

1.9375

1.875

1.9375

15

T: 1m/s

A: 2m/s

T: 1m

1.75

1.875

1.9375

1.5

0

1

T: 0.5m

A: 1m

1.75

1.875

1.9375

1.5

0

1

A: 0.5m

T: 0.25m

1.75

1.875

1.9375

1.5

0

1

I could continue this but I would never reach 2 metres, the point where Achilles is set to overtake the tortoise, since if you continue adding fractions with the next power of two as the denominator and 1 as the numerator to the number one, (e.g. 1 + 1 2 + 1 4 + 1 8 … ) you will never reach two. What would happen, however, if the tortoise travelled at 0.01m/s? Achilles would still never catch up! What if the tortoise travelled at 0.001m/s? Achilles could not catch up. What if the tortoise travelled at 0.0…01m/s? The tortoise is travelling so slowly it is not moving, yet Achilles still cannot catch up, therefore, taken to an extreme, this paradox states moving isn’t possible. As movement is one of the seven life processes making a being living, in the limit, this paradox states that nothing is alive! Therefore, Zeno is correct in a sense, however, as demonstrated by my example, the proof of his paradox has a limit. In my example, the limit is 2m. So, what he paradox is stating is that, given the starting speeds, it is impossible for Achilles to overtake the tortoise in 2 metres. However, if you extend the race beyond this limit, it becomes untrue and it has been proven many times that races can be over the limit, for example in the Olympics. Zeno only considers a part of reality by considering only the section of the race before the limit, so Zeno’s paradox is true within the limit, but, as I have illustrated, untrue if the limit is breached. Despite this, this paradox did lead us to realise that finite things can be split infinite times. The concept of an infinite series is used today in finance to work out mortgage payments, which is why they take an infinite amount of time to pay off! So, thanks to Zeno, I apparently must live to an age of 167 just to afford a garden shed! This is a valid way to structure the problem, but it means that Achilles will never reach the point where he overtakes the tortoise, and a conclusion is that we are all actually dead, oh dear.

16

Parity

Hin Chi Lee

Parity can be defined as whether the integer being odd or even, and the relationship between them. Here are the following rules all integers always follow:

Even + Even = Even eg 2 + 2 = 4

Even + Odd = Odd eg. 2 + 3 = 5

Odd + Odd = Even eg. 1 + 3 = 4

Even x Even = Even eg. 2 x 2 = 4

Even x Odd = Even eg. 2 x 3 = 6

Odd x Odd = Odd eg. 3x 5 = 15

The concept of parity can be quite useful particularly in disproving an argument that is based on two variables. We take the parity of an odd number as 1, while the parity of an even number as 0. Parity can also be treated as a modulo(2) operation or binary system. When two odd numbers of parity 1 are added together, they have a parity of 2, which cancels to become 0, hence the sum of two odd numbers is an even number.

𝑃𝑖 () + 𝑃𝑖 () = 𝑃𝑖 ( + )

Here are examples which can be solved by parity rules and arguments.

John and Sam are playing are tearing paper after an exam. They start off with 3 pieces of paper, and every time John picks up a piece of paper, he tears that piece of paper into 3 smaller pieces. In a similar fashion, Sam tears the paper he picks up into 5 smaller pieces. They can tear the paper in any order and however many times they like (eg. John can tear it 10 times while Sam doesn’t do it at all). So the question is: is it possible for there to be a total of 100 pieces of paper in any scenario? Have a think. Answer is on the following page.

1

2

5

1

3

3

4

2

John – tears into 3 pieces

Sam – tears into 5 pieces

17

Here, if we take the original number of pieces of paper as n. We can see that every time John tears the paper, it adds 2 pieces to the total count (n+2), while Sam adds 4 pieces to the count (n+4). Both of these additions do not change the parity of the number, which in this case, the total number of pieces of paper always stays as an odd number . Hence we can deduce it is impossible to have 100 pieces of paper.

Again if we had the same rules but the game has started off with 4 pieces of paper, would it be possible to have a total of 100 pieces of paper? Have a think about it.

Let’s have another example. If we take an 8x8 checkerboard, and we want to cover it with by a 1 x 2 domino, it is not hard to see that we can cover up the whole checkerboard full. If we take out the two corner squares, it is possible to cover up the remaining squares?

Use these to fill up the whole board!

With parity arguments, we can disprove this possibility. Since every adjacent square of a black square is a white square, the 1 x 2 domino must cover 1 white square and 1 black square. If we count up the number of black squares and white squares, we can see that there are a total of 32 black squares and 30 white squares. It is impossible to pair up all the black squares with enough white squares. Thus, it is impossible to fill up the whole area with 1 x 2 dominos.

Fun fact: The word parity comes from the pairing and equivalence of things.

There is also another type of parity: the parity between positive and negative numbers. When two negative numbers are multiplied, you get a positive number; when you multiply a positive and a negative, the result is a negative number; and lastly when you multiply two positive numbers, you get a positive numbers. This is common sense – but we can observe these positive and negative numbers follow a similar pattern as the odd-even parity. We can use these rules to solve the problem below:

18

Given that 1

· 2

· 3

… · 30

= 1 , and all of the numbers are integers,

Is it possible for 1

+ 2

+ 3

+ ⋯+ 30

= 0 ?

We observe that all the numbers must be any integers, so we can deduce that all numbers must be either +1 or -1, since for example, having a number like +2 would require a -1/2 as a pair to create the +1 in the answer for the first equation. In order for the second equation to be possible, we must have the same number of +1s and -1s so that together they give an answer of 0. This means we have in total fifteen pairs of +1s and -1s. If we try to approach the first equation, since we know that the product of a pair (+1 × −1) gives -1 with the parity rules, the product of these fourteen pairs must give +1. Wait, there’s the 15 th pair left! So +1 x -1 gives an answer of -1, which isn’t the answer we are seeking from the first equation. We can thus conclude that these two equations are impossible having used rules of parity.

Now let’s look at another scenario below:

One day, you and nine other friends are walking down the street and an UFO secretly lands beside you. All 10 of you are captured by these aliens and now they put you to a test to investigate whether humans are intelligent life forms. If you succeed, you will be safely released back to Earth.

They will put you into a row of 10 in descending heights and facing the front, so that the tallest person at the back can see everyone in front, and the second tallest person can see the 8 people in front of him etc. Then, either black or white hats are put on everyone in a completely random fashion, and you can only see the colour of the hat of the people in front of you. In order for all of you to be released, at least 9 people have to guess correctly of the colour of their own hat. The order of the guesses has to be

from the tallest person to the shortest. Unfortunately, if you say anything other than “black” or “white”, or try to do anything suspicious by changing the tone of your voice or other actions, you will be fried instantly by their laser gun. Everyone can hear the guesses of all the previous people. Now, before you are put to the test, you all are allowed 10 minutes to discuss and form a plan. Can you overcome this test? 1 Firstly, we observe that there are only two possible colours of hats, which means it is possible we can use ideas of parity to solve this, either a white hat, or a non-white hat (black). The method is as follows: the tallest person establishes the parity of the number of white hats. So he could say “black” to suggest that he sees an odd number of white hats, or say “white” if he sees an even number. He has a 50% of guessing the colour of his own hat correctly, but this doesn’t matter – as long as the other 9 gets it right. Using the above picture as an example, the tallest person will say “white” as he sees 6 people with white hats. This tells the second tallest person that he sees an even number of white hats, but the second person can only see an odd number of 5 white hats in front of her. The second tallest person

1 https://www.youtube.com/watch?v=N5vJSNXPEwA&vl=en – TED talk by Alex Gendler

19

observes a switch in parity (from even to odd), meaning that she herself is wearing a white hat which causes the change in parity! So she says “white” the correct answer. The third tallest person now knows that the parity of white hats that he should observe now an odd number having heard from the first two people. He observes 5 white hats, an odd number, meaning that there is no change in parity, thus he is able to deduce that he is wearing a black hat. This process repeats again until the shortest person successfully guesses the correct answer. Congratulations! All of you are safely released! Looks like humans are intelligent life forms after all.

20

Taylor Series and Euler’s identity

Andrew Ng

The square root of -1 is denoted by the letter 𝑖 . 𝜋 is the ratio of the circumference of a circle to its diameter, and e is defined as the constant satisfying the following limit:



1 

lim →∞

(1 +

)

The latter two can be shown to be both irrational (i.e. they can’t be written in the form  where and  are integers) and transcendental, i.e. they aren’t the solution of a polynomial equation with rational coefficients, e.g. x-1=0. What would go through your mind if I told you that:

 𝜋 = −1

Mind-blowing eh? From this 𝑖 can be shown to equal  − 𝜋 2 It seems completely inconceivable to the uninitiated that raising an infinitely long constant to another infinitely long constant multiplied by an imaginary number gives such a normal integer as -1.

It all arises from the following formula:

 () () !

( − ) 

() = ∑

=0

where  () () denotes the  th derivative of () evaluated at the point .

Intuitive Proof:

Given any function () that is defined and infinitely differentiable over [−∞, ∞] , we can approximate it as a polynomial which we call the power series of () . Let’s call it 𝑃() . It must then be equal to () at at least one point,  . Then, let:

( − ) 2 + ⋯

𝑃() = 0

+ 1

( − ) + 2

Evaluating at  , () = 0

. Therefore:

( − ) 2 + ⋯

𝑃() = () + 1

( − ) + 2

Differentiating 𝑃() :

( − ) 2

𝑃’() = 1

+ 2

( − ) + 3

Repeating the above process, we have

1

= ’() .

Differentiating P’(x) and repeating the above process, we have

21

2 2

= ’’()

′′() 2!

2

=

Since 2=2! Differentiating and repeating the above process infinitely many times, we have:

3() 3!

4() 4!

3

=

4

=

Since all derivatives of () are identical to those of 𝑃() ; the curves change in an identical way and, since they start from the same point, the curves are identical, i.e.

() 0!

′()( − ) 1!

′′()( − )2 2!

() ≡ 𝑃() ≡

+

+

Condensing this infinite sum into sigma notation yields the formula given in the introduction.

Examples:

Below are examples of the Taylor series centred at the point p=0 for common functions. Taylor series centred at the point p=0 are also known as Maclaurin series.

2 2!

3 3!

1!

 𝑥 = 1 +

+

+

. . . . .

2 2!

4 4!

6 6!

() = 1 −

+

. . . .

3 3!

5 5!

7 7!

𝑖() = −

+

+ ⋯

Placing any value for x into the right hand side polynomial gives an approximation for the actual function on the left hand side of the equation. For example, putting = 0 into the trigonometric functions gives the standard results cos(0) = 1 and 𝑖(0) = 0 .

The series also works if composite functions are used. For example,

(2) 2 2!

(2) 3 3!

2 1!

 2𝑥 = 1 +

+

+

Some properties of the imaginary number 𝑖 include:

𝑖 1 = 𝑖 , 𝑖 2 = −1 , 𝑖 3 = −𝑖 , 𝑖 4 = 1 …

Generalising to all integers:

𝑖 4+1 = 𝑖 , 𝑖 4+2 = −1 , 𝑖 4+3 = −𝑖 , 𝑖 4 = 1

where n is a non-negative integer.

Replacing with 𝑖 in the exponent of  𝑥 gives:

22

(𝑖) 2 2!

(𝑖) 3 3!

(𝑖) 4 4!

𝑖 1!

 𝑥 = 1 +

+

+

+

. . .

Notice that this becomes:

2 2!

𝑖 3 3!

4 4!

𝑖 5 5!

𝑖 1!

 𝑥 = 1 +

+

+

. . .

Grouping the terms gives:

2 2!

4 4!

6 6!

3 3!

5 5!

(1 −

+

. . . ) + 𝑖( −

+

. . . )

However, since the first parenthesis is the Taylor series for cos(x) and the second parenthesis is the Taylor series for sin(x), we can transform the equation into:

 𝑥 = () + sin()𝑖

Finally, evaluating at 𝜋 :

 𝜋 = −1

which is Euler’s identity.

Applications of Taylor Series

Apart from the unbelievable result obtained above, there are many other applications for this formula. For example, there is no way to evaluate the indefinite integral ∫ 𝑖( 2 ) via conventional means. However, by substituting 2 =  into the equation for 𝑖() and integrating each of the individual terms a reasonably accurate approximation can be found. On a similar note, if you ever find yourself having forgotten your calculator in an exam you can find a numerical value to an appropriate degree of accuracy for certain functions using a Taylor series (this is definitely not recommended). For those of you who are fortunate (or unfortunate) enough to have to solve differential equations on a regular basis, the Taylor series can be a weapon to add to your repertoire. For example: ’’ + 2 =  𝑥 . This differential equation is not at all trivial to solve, but by representing y as a Taylor series it’s possible to approximate a solution. Further uses include: evaluating infinite sums, generating functions in combinatorics, evaluating energy functions in physical systems at equilibrium, power flow analysis and many more.

Challenge:

Bibliography:

Can you prove the result

i i =e - π/2

http://math.stackexchange.com/questions/218421/what- are-the-practical-applications-of-the-taylor-series

as given above?

23

What is the Probability of Two Random Numbers Being Coprime?

Theo Macklin

If one were to select two integers at random between one and infinity, what is the likelihood that they would share no factors other than 1? This seems like a question that would venture into the realm of prime numbers or other unresolved questions. However, this question actually leads us along a path through a wide variety of mathematical fields and uses one of the most compelling results in mathematics. To start to find an answer for this question we must first phrase our question in terms of its mathematical operators. To do this we must appreciate that when two numbers are coprime their greatest common denominator is 1. For two integers, and  , this expression is as follows:

gcd[, ] = 1

The gcd function outputs the greatest common factor of its two operators. We can now use probability notation to express the probability of such a state occurring out of 1 (i.e.: 50% chance ≡ 0.5). We will call this probability  :

𝑃(gcd[, ] = 1) = 

Let this be result (1)

Now let us consider the alternative possibility: a number 𝑘 is the greatest common denominator of two numbers, and b , where ( , b) > 𝑘 . The likelihood that both numbers are both divisible by 𝑘 is equal to the probability of each number being divisible by 𝑘 multiplied together. Since it is logical to assume that 2 divides half of all numbers, 3 a third, 4 a quarter, 5 a fifth, etc. it is also reasonable to conclude that 𝑘 divides one 𝑘 th of all numbers. Thus:

𝑃(𝑘 𝑖   ℎ  ) = 𝑃 ( 𝑘

 𝑘

∈ ) × 𝑃 (

∈ )

1 𝑘

1 𝑘

1 𝑘 2

=

×

=

To determine whether 𝑘 is the greatest common denominator of and b we must ensure that:

𝑘

 𝑘

gcd [

,

] = 1

As 

≡ and  

≡  the probability that they are coprime must be, as in result (1) , y .

𝑘

 𝑘

𝑃(gcd [

,

] = 1) = 

Thus, the probability that 𝑘 is the highest common factor must be:

24

𝑃(𝑘 𝑖   ℎ  ) × 𝑃(gcd [ 𝑘 ,

 𝑘

] = 1)

1 𝑘 2

 𝑘 2

×  =

Using the knowledge that two numbers must have a greatest common denominator, we know that the sum of the probabilities where 𝑘 = 1, 2 ,3, … , ∞ must cover all pairings and so equal 1:

 𝑘 2

= 1

=1

Rearranged:

1 𝑘 2

∑

= 1

=1

1

 =

1 𝑘 2

∞ =1

Let this be result (2)

Euler’s Solution to the Basel Problem

The result for the sum to infinity of 𝑘 −2 is called the Basel problem. It was investigated in the 17 th and 18 th centuries until its exact solution was proved in 1734 by Euler. His initial proof (below) was criticised due to his assumption that what is true for a finite polynomial is also true for an infinite one. Despite this criticism, Euler later went on to prove that his result had been correct. Ever since, this result has been widely used and has been proved in countless other ways; however, Euler’s is still the most elegant solution.

To commence the proof, Euler considered the Maclaurin series of sin - the expression that equals the function but represented in powers of . The Maclaurin series takes the general form:

2  ′′ (0) 2!

3  3 (0) 3!

′(0) 1!

() = (0) +

+

+

+ ⋯

Applied to () = sin :

− 2 sin(0) 2!

− 3 cos(0) 3!

4 sin(0) 4!

5 cos(0) 5!

cos(0) 1!

sin = sin(0) +

+

+

+

+

+ ⋯

3 3!

5 5!

sin = 0 + − 0 −

+ 0 +

+ ⋯

3 3!

5 5!

7 7!

2+1 (2𝑘 + 1)!

+ ⋯+ (−1) 

sin = −

+

25

This is the type of series that is used by calculators to accurately approximate trigonometric ratios. Euler then considered a finite  th degree polynomial, p(x) , with two properties:

1. 𝒑() has non zero roots :  

,  

, … ,  𝒏

2. 𝒑() = 

Such a polynomial can take the form:

1

2



() = (1 −

) (1 −

) × …× (1 −

)

Euler continued:

2 3!

4 5!

6 7!

sin

= 1 −

+

+ ⋯

Knowing that the roots, where () = 0 , of sin occur at integer multiples of 𝜋 (if you have remembered to put your calculator into radians) allows us to represent the above series as follows:

sin

𝜋

𝜋

2𝜋

2𝜋

= (1 −

) (1 +

) (1 −

) (1 +

) × …

2 𝜋 2

2 4𝜋 2

2 9𝜋 2

sin

= [1 −

] [1 −

] [1 −

] × …

sin𝑥 𝑥

is an infinite series. That it can be represented like a finite polynomial was Euler’s underlying assumption in this proof. It was the subsequent definitive proof of this element using the Weierstrauss Factorisation Theorem that led to the result being widely accepted as correct.

Expanding the above product gives:

sin

1 𝜋 2

1 4𝜋 2

1 9𝜋 2

= 1 − 2 (

+

+

+ ⋯) + ⋯

2 3!

4 5!

6 7!

1 𝜋 2

1 4𝜋 2

1 9𝜋 2

+ ⋯ = 1 − 2 (

1 −

+

+

+

+ ⋯) + ⋯

Simplifying both sides to compare the 2 terms:

2 3!

1 𝜋 2

1 4𝜋 2

1 9𝜋 2

= − 2 (

+

+

+ ⋯)

1 6

1 𝜋 2

1 4𝜋 2

1 9𝜋 2

=

+

+

+ ⋯

𝜋 2 6

1 2 2

1 3 2

1 4 2

= 1 +

+

+

+ ⋯

Thus:

𝝅  

 𝒌 

=

𝒌=

26

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